Correct, but it is pertinent to note that as the frequency decreases below 50Hz the torque will stay roughly constant, unlike a belt or gear change arrangement. So the motor output power will go down in proportion to the frequency decrease.

Andrew

This might seem a bit of a cack-handed reply, as I can't answer your specific question from an electrical point of view – I'm not qualified to do so. What I can do, is tell you of my practical experience of doing a similar motor swap on my Perfecto shaper.

As part of my lock down boredom relief, I decided to re-motor my shaper and lathe with 3-phase motors and VFDs. The lathe was going to be uprated to 1HP from the standard ¾HP as I considered it lacked sufficient grunt. The shaper, on the other hand was powered by a ⅓ HP single phase motor. I found a replacement 3-phase motor of the same power rating but not a matching VFD. I subsequently purchased a super-cheap 0.45kW VFD online and fitted that. It works fine with no apparent ill effects on either piece of equipment.

There is a short video of the set up in action here: Shaper in action.

So, from a purely practical point of view, I'd say "Go ahead."

John

2

g head."

If you plan to run the motor much below normal speed for extended periods, consider motor cooling.

The built in motor fan is less effective at lower motor speed.

The VFD may deliberately limit it's output to effectively de-rate the motor when speed drops in order to protect the windings (that might be a user selectable option), it's also possible to add an external fan to the motor to force cool it regardless of speed.

Oh dear, that's very disappointing, I thought SoD was smarter than that. He should go and stand in the corner and contemplate his misdemeanor!

In a simple model of an induction motor torque is proportional to current. As the frequency, and speed, decrease from base speed the VFD will decrease the applied voltage to keep the current constant. So the torque is essentially constant below base speed. For the purposes of this thread we'll take base speed as that when the motor is running from a 50Hz supply. As the motor speed drops the torque stays constant but the output power drops in proportion. Gear and pulley arrangements increase the torque in proportion to the speed reduction so the power stays constant.

Newer VFDs, that use vector control, can increase the current, and hence torque, below base speed. To some extent that ameliorates the drop in output power. But it is normally intended to counteract short term overload, not for continuous operation. The motor is being run beyond its design current and will quickly overheat, even with a fan running at normal speed.

I'd agree with the above that a larger VFD will have no problem running a lower power motor. The only real problem occurs when the VFD is capable of much more power than the motor needs. The internal current sensors will be sized according to the rated motor power. If a smaller motor is used naturally the currents will be lower and the sensor will be operating over a small part of it's range, leading to a loss of resolution. That may be a problem for the more complex control algorithms.

I'm not sure why a belt sander needs variable speed control?

Andrew

There is a short video of the set up in action here: Shaper in action.

Nothing to do with the original question, but a shaper use query from the video : why are you incrementing the traverse on the cut stroke ?

I usually set my Boxford to do the feed increment on the return stroke, that way the cut is perpendicular to the traverse – your cut is at an angle, as can be seen at the end of the pass on the block where the last bit to be cleared isn't parallel to the edge.

I appreciate that for the operation shown it doesn't matter, but for shaping up to a shoulder it would. Not intending to be nitpicky, just curious.

Nigel B.

Edited By mgnbuk on 13/02/2021 14:29:13

Wot me make a mistake on the forum? Surely not!

In this case yer honour I was thinking of this graph:

But Andrew's quite right, I should have explained the context. And maybe the graph doesn't cover how a 3-ph motor behaves when it's speed is reduced by altering frequency! It's another pause for thought moment.

What I think I know doesn't help much. How different types of electric motor behave during start up and low speed varies. A single-phase motor suffers low torque until it builds up speed, but a stepper motor has high torque even when stopped. I'd be in trouble if asked to program the output voltage, frequency and current of a VFD because none of my books describe how 3 phase motors behave at anything other than 50 or 60Hz. (Or if they do, I don't understand it.)

Here's a question. Is there an easy way of measuring the torque available at a lathe's chuck at various speeds? If so would be interesting for members to see what their treadle, single-phase, DC-PWM, 3-phase, VFD, or brushless motor is actually delivering.

Dave

My Milling machine came with a 1/2 hp motor fitted with a Toshiba 0.75kw inverter. All has been well since owning it & not had any torque issues, until.

I got hold of a auto tapping head & was doing some tests with it. I was running the inverter at 12hz, rather than changing belt speeds. At that frequency the torque was pretty low. I could not even drive a 6mm tap through 10mm alloy plate.

On my lathe i have a Omron inverter with a 1.5hp motor. I often run this at 1.5 hz when putting glue on rings. It runs happily at that frequency & seems to have bags of torque. That inverter has a setting called torque boost, which maybe explains the power. + the fact the motor is much bigger than on the mill. Or was. I have just changed the Mill motor for a 1 hp. Not done any more tests as yet.

In answer to your question , yes it will work fine. As others have already said, go through the settings & match the motor current .

Steve.

I can see a few lathes with bruised beds coming up.

Steve.

Like the title says the graph shows the torque-speed curve of a 3-phase motor. What it doesn't say is that it is at a fixed frequency. Let's assume that we have a 3-phase motor running off a 3-phase supply with no load and no internal losses. The motor will run at the synchronous speed, for a 4-pole motor and 50Hz that is 1500rpm. At this point we are at the bottom right of the graph. Of course a practical motor will have some internal mechanical losses which require torque to overcome. To provide this torque the motor has to slow down below synchronous speed; it slips below synchrnous speed. So we move up the torque curve with a slight decrease in speed. If we now add a mechanical load to the motor then we continue up the curve. The speed drops more and the torque produced rises as the current rises. When we reach the rated power of the motor the current will be the rated current and the speed will have dropped to the rating plate value. For most motors the speed at rated power is a few percent less than the synchronous speed.

If we now continue to increase the load the speed will drop further, the currents will rise above rated current and the torque will also rise. At some point we reach the maximum torque – pull out torque. This is normally 250% or so more than rated torque and interestingly is determined by rotor resistance. A VFD can mimic this behaviour in a controlled manner by trading speed for increased torque during a short term overload. My CNC mill does this.

If the motor load continues to rise the speed drops further and the torque also drops. Of course from full rated power onwards the motor is overloaded and will probably have thermal issues.

To summarise, the graph is nothing to do with how the torque varies with frequency for a motor driven by a VFD and within it's rated power.

Measuring torque on a rotating shaft is involved, and done properly requires expensive inline sensors and modification of the drive train. Less accurate estimates of the torque in a rotating shaft can be made by measuring the twist of the shaft under load.

Andrew

Two quick replies:

Emgee: Thanks for the comment. I bought the shaper some time ago, on a whim as it was available at the right price and I only had room for one that size. I don't suppose I've used it more than a couple of times in the last three years and only then to try it out.

mgnbuk : I'm afraid I don't understand to what you refer. The shaper cuts on the forward stroke and the traverse ratchets round on the return, ready for the next cut. That's how I understood it to work. There doesn't appear to be any way to alter that arrangement.

And yes, I agree, my setting up wasn't totally square and it cuts a bit on the skew, but it was only supposed to be a test of the effect of the different speeds to be sure I hadn't completely wasted my dosh on the 3-phase set up, really. I'll try to do better, next time!

John

Edited By John Hinkley on 13/02/2021 17:03:39

Nigel,

I've had another look at the video and now I can see what you describe. It shouldn't do that! I'm now wondering if a previous owner has dismantled it and put it back together incorrectly, if that is possible? Tomorrow, I'll investigate as it's forecast to get a little warmer.

Thanks for pointing it out. It might also explain the squinty cut, too! Maybe my set up wasn't that bad , after all.

John